Question 7 of 7: Reading the Ellingham Diagram — Ti/TiO₂ Equilibria

Question 7: Reading the Ellingham Diagram — Ti/TiO₂ Equilibria (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The attached Ellingham diagram (Fig. 9-3, Gaskell) plots $\Delta G^\circ=RT\ln P_{O_2}$ for oxide-formation reactions against temperature, with nomographic $\text{CO/CO}_2$ and $\text{H}_2/\text{H}_2\text{O}$ scales.

Find. (a) $P_{O_2}$ at the Ti/TiO₂ line at 1500 °C; (b) the CO/CO₂ ratio giving the same oxygen potential at 1000 °C; (c) the H₂/H₂O ratio at 1000 °C; (d) $\Delta G^\circ$ for Ti + SiO₂ = TiO₂ + Si at 1000 °C; (e) why the lines have no vertical jump at a metal's melting/boiling point.

Approach. Build the straight-line $\Delta G^\circ(T)$ for Ti+O₂=TiO₂, 2CO+O₂=2CO₂, 2H₂+O₂=2H₂O and Si+O₂=SiO₂ from 298 K enthalpy/entropy data (exactly what the diagram encodes). Part (a) reads $P_{O_2}$ straight off the Ti/TiO₂ line; parts (b)/(c) equate that same oxygen potential to the potential fixed by a CO/CO₂ or H₂/H₂O ratio; part (d) subtracts the Si/SiO₂ line from the Ti/TiO₂ line (equivalent to reading the vertical gap between the two lines).

1. Ti/TiO₂ line. With $\Delta H^\circ_f(\text{TiO}_2)=-944{,}000$ J/mol, $S^\circ(\text{TiO}_2)=50.6$, $S^\circ(\text{Ti})=30.7$, $S^\circ(\text{O}_2)=205.1$ J/mol·K: $\Delta S^\circ=50.6-30.7-205.1=-185.2$ J/mol·K, so $$\Delta G^\circ_{Ti}(T)=-944{,}000+185.2\,T\ \text{J/mol}.$$
2. (a) $P_{O_2}$ at 1500 °C ($T=1773.15$ K). $\Delta G^\circ_{Ti}(1773.15)=-944{,}000+185.2(1773.15)=-615{,}600\ \text{J/mol}$. Since $\Delta G^\circ=RT\ln P_{O_2}$, $$P_{O_2}=\exp\!\left(\frac{\Delta G^\circ_{Ti}}{RT}\right)=\exp\!\left(\frac{-615{,}600}{(8.314)(1773.15)}\right)=\boxed{7.3\times10^{-19}\ \text{atm}}.$$
3. (b) CO/CO₂ ratio at 1000 °C ($T=1273.15$ K). Build the $2\text{CO}+\text{O}_2=2\text{CO}_2$ line the same way ($\Delta H^\circ=-566{,}000$ J, $\Delta S^\circ=-172.8$ J/K): $\Delta G^\circ_{2CO}(1273.15)=-345{,}900$ J. Matching oxygen potentials, $\left(\dfrac{P_{CO_2}}{P_{CO}}\right)^2\exp\!\big(\Delta G^\circ_{2CO}/RT\big)=\exp\!\big(\Delta G^\circ_{Ti}/RT\big)$, so $$\frac{P_{CO}}{P_{CO_2}}=\exp\!\left(\frac{\Delta G^\circ_{2CO}-\Delta G^\circ_{Ti}}{2RT}\right)=\exp\!\left(\frac{-345{,}900-(-708{,}200)}{2(8.314)(1273.15)}\right)=\boxed{2.7\times10^{7}}.$$ A huge excess of CO over CO₂ is needed — TiO₂ is far more stable than CO₂, so an extremely reducing atmosphere is required before Ti stays un-oxidized.
4. (c) H₂/H₂O ratio at 1000 °C. The $2\text{H}_2+\text{O}_2=2\text{H}_2\text{O}$ line ($\Delta H^\circ=-483{,}600$ J, $\Delta S^\circ=-88.8$ J/K) gives $\Delta G^\circ_{2H_2}(1273.15)=-370{,}600$ J, so by the same construction $$\frac{P_{H_2}}{P_{H_2O}}=\exp\!\left(\frac{-370{,}600-(-708{,}200)}{2(8.314)(1273.15)}\right)=\boxed{8.4\times10^{6}}.$$
5. (d) $\Delta G^\circ$ for Ti + SiO₂ = TiO₂ + Si at 1000 °C. This reaction is (Ti+O₂=TiO₂) minus (Si+O₂=SiO₂), so $\Delta G^\circ_{rxn}=\Delta G^\circ_{Ti}-\Delta G^\circ_{Si}$. With $\Delta H^\circ_f(\text{SiO}_2)=-910{,}900$ J, $S^\circ(\text{SiO}_2)=41.5$, $S^\circ(\text{Si})=18.81$ J/mol·K: $\Delta G^\circ_{Si}(1273.15)=-678{,}700$ J, so $$\Delta G^\circ_{rxn}=-708{,}200-(-678{,}700)=\boxed{-29.5\ \text{kJ/mol}}.$$ Negative, so titanium spontaneously reduces silica at 1000 °C — consistent with the Ti line sitting below the Si line on the diagram.
6. (e) Why no discontinuity at a phase transition. At a metal's melting or boiling point $T_{tr}$, the solid and liquid (or liquid and vapour) phases are, by the definition of that equilibrium temperature, at equal Gibbs energy. The oxidation-reaction $\Delta G^\circ(T)$ line is therefore continuous straight through $T_{tr}$ — it cannot jump, because both phases give the same value at that one temperature. What does change discontinuously is the reaction entropy $\Delta S^\circ$ (the metal gains entropy on melting or boiling, while the oxide's entropy is essentially unaffected), and since the line's slope is $-\Delta S^\circ$, the line bends — a change in slope (a kink), never a vertical jump in value.